Consider the following position function. Find (a) the velocity and the speed of the object and (b) the acceleration of the object. r(t) = (9,t^3, e^-t), for t greater than or equal to 0

v = 3t^2, -e^-t ;

Speed = √ 9t^4 + (-e^-t)^2 ;

a(t) = 6t, e^-t

Step-by-step explanation:

Given the position function at time 't' for t greater than or equal to 0:

r(t) = (9,t^3, e^-t)

The Velocity of 'v' is the first derivative of r with respect to 't' :

d/dt(9) = 0

d/dt(t^3) = 3t^3-1 = 3t^2

y = (e^-t) =

Using chain rule :

Let u = -t

y = e^u ; dy/du = e^u

du/dt = -1 ;

Therefore dy/dt = dy/du * du/dt

dy/dt = -1 * e^u ; -e^u ; u = -t

dy/dt = -e^-t

Therefore,

v = 3t^2, -e^-t

The magnitude of velocity 'v ' = speed

Speed = √v1^2 + v2^2

Speed = √(3t^2)^2 + (-e^-t)^2

Speed = √ 9t^4 + (-e^-t)^2

Acceleration(a) = derivative of velocity

d/dt(3t^2) = 6t

d/dt (-e^-t) :

Using chain rule :

dy/dt = -e^-t

Let u = -t ; du/dt = -1

-e^u ; dy/du = -e^u

Therefore, dy/dt = -1 * -e^u

u = -t

dy/dt = e^-t

Therefore, acceleration:

a(t) = 6t, e^-t

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too hard the other guy did it

step-by-step explanation: